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Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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Is m + z > 0? (1) m  3z > 0 (2) 4z  m > 0
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Originally posted by kman on 11 Feb 2009, 16:20.
Last edited by Bunuel on 26 Apr 2018, 01:47, edited 2 times in total.
Added the OA.




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Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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16 Dec 2010, 09:29




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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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Updated on: 26 May 2011, 11:13
Plus in arbitary values to check is m+z is always greater than 0 or there are other possibilities
st1 ==> m  3z >0 m =1 and z = 1 ==> m+z = 0 m=2 and z = 1 ==> m+z >0 not sufficient
st2 ==> 4zm>0 m = 1 and z = 0.1 ==> m+z < 0 m=2 and z = 1 ==> m+z >0 not sufficient
combining both
m3z + 4z m >0 z > 0.... (P) from 1, m > 3z and from 2, 4z > m ==> 3z < m < 4z... (Q) from (P) and (Q) m+Z > 0
Answer C.
Originally posted by agdimple333 on 26 May 2011, 10:10.
Last edited by agdimple333 on 26 May 2011, 11:13, edited 1 time in total.




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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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11 Feb 2009, 19:57
I will go with C.
From clue 1, m3z > 0 here m, z can be positive or negative. So cannot say whether m + z > 0. Hence Insufficient.
From clue 2, 4zm > 0, here m, z can be positive or negative. So cannot say whether m + z > 0. Hence Insufficient.
Cosider both the clues.
m3z>0>Ine1 4zm>0 >ine2
Add both the inequalities, z > 0. m3z > 0 ==> m > 3z. Since Z is positive m also should be positive.
So we can say that m+z > o. Hence sufficient.



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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12 Feb 2009, 02:38
It is also possible to solve it by drawing: 1. Let's draw our statement using two colors: red (statement is false) and green (statement is true). We will have two color cake: Attachment:
t75657_1.png [ 6.27 KiB  Viewed 24020 times ]
2. Now, cut a bit of the cake by first condition: Attachment:
t75657_2.png [ 9.33 KiB  Viewed 24006 times ]
3. cut a bit of the cake by second condition: Attachment:
t75657_3.png [ 8.99 KiB  Viewed 24000 times ]
4. Now, use both conditions: Attachment:
t75657_4.png [ 10.54 KiB  Viewed 24008 times ]
So, C is obvious after drawing as the statement is only true.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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17 Dec 2010, 09:13
smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. Bunuel, is m+z > 0 the same as m/z > 1 ? if so, then 1. would be m/z > 3, which is SUFF and 2. would be m/z < 4, which is INSUFF so I would have said A Please correct me if I'm wrong  my inequality skills are a bit rusty



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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17 Dec 2010, 09:25
psirus wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. Bunuel, is m+z > 0 the same as m/z > 1 ? if so, then 1. would be m/z > 3, which is SUFF and 2. would be m/z < 4, which is INSUFF so I would have said A Please correct me if I'm wrong  my inequality skills are a bit rusty No, it's not correct. When you are writing m/z>1 from m+z>0 you are actually dividing both parts of inequality by z: never multiply or reduce (divide) inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.So if we knew that z>0 then m+z>0 > m/z+1 >0 and if we knew that z<0 then m+z>0 > m/z+1 <0. Hope it's clear.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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19 Jan 2011, 22:14
Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. Hello Bunnel, As you said in the above post.. "(1)+(2) Remember we can add inequalities with the sign in the same direction >"what should we have different signs...



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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20 Jan 2011, 02:19
jullysabat wrote: Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m3z > 0 2. 4zm > 0
OA : C
Need help in underdstanding how we arrive at C. (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. Hello Bunnel, As you said in the above post.. "(1)+(2) Remember we can add inequalities with the sign in the same direction >"what should we have different signs... You can only add inequalities when their signs are in the same direction:If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) > \(a+c>b+d\). Example: \(3<4\) and \(2<5\) > \(3+2<4+5\). You can only apply subtraction when their signs are in the opposite directions:If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\).
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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21 Nov 2012, 19:34
arvindbhat1887 wrote: Is m+z>0?
(1) m > 3z (2) m < 4z.
The answer is given as (C) Both statements together. I don't understand how. Can someone please explain? I think the answer is (E) Not sufficient with both (1) tells you that m > 3z (2) tells you that m < 4z. Either of the two cases taken individually is not sufficient. So Rule out (A), (B), (D) Therefore, 3z<m<4z when you combine the two. Now z can take both ve and +ve values. So, m + z can be either ve or +ve depending on the value of z. Hence, (E). Please help. Stem 1: m>3z or m3z >0 doesnt tell us anything about m+z. Not sufficient. Stem 2: m<4z or m4z <0 or 4zm >0 tells us nothing again. Not sufficient. combining, we get that 3z <m <4z also, if we add both equations, m3z >0 4zm >0 we get, z>0 Thus since m>3z => m is also >0 therefore m+z >0 Sufficient. Ans C it is.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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21 Jul 2013, 01:29
Bunuel wrote: (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlI have one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z  m > 0 So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately?



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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21 Jul 2013, 01:37
fozzzy wrote: Bunuel wrote: (1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction > \(m3z+4zm>0\) > \(z>0\), so \(z\) is positive. From (1) \(m>3z=positive\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive) > \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: ismz01m3z024zm75657.htmlI have one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z  m > 0 So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately? You got that m is positive with (1), so can stop there. If you combine you get that 4z>m>3z>0.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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07 Sep 2014, 07:41
Hi Bunuel, I have a doubt in this question. it is asked whether m+z>0 this implies that is m>z?
Statement 1 says: m3Z>0 then m>3z and 3z is definitely greater than z then isnt m>z. Please clarify.



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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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23 Dec 2015, 22:28
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is m+z > 0 (1) m3z > 0 (2) 4zm > 0 In the original condition, there are 2 variables(m,z), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1)&2), add the 2 equations, which is m3z+4z+m>0, z>0. Then m>z>0 is derived from m>3z>z, which is m>0. So, m+z>0 is always yes and sufficient. Therefore, the answer is C. > For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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09 May 2016, 20:55
smitakokne wrote: Is m+z > 0
(1) m3z > 0 (2) 4zm > 0 Responding to a pm: Quote: When I look at such question, I know I will figure out something if I consider both statements together. I feel comfortable with applying inequalities rules there. But I find it harder to approach individual statements in such cases. I get lost in figuring out cases to reject (or approve) one statement on its own. At times, I take time and eventually rush onto considering both statements together.
Can you please help me out if I'm lacking some approach or concepts?
There really isn't one suitable method for all inequality questions. The approach depends on the kind of question. Very rarely will I plug in numbers here to find cases where the inequality holds or does not hold. This is how I will do this question: "Is m+z > 0?" Here I think that m and z could both be positive or one of them could be positive with greater absolute value than the other. If both are negative, this doesn't hold. Go on to stmnts. (1) m3z > 0 Here, I will try to segregate the variables to get relation between them. m > 3z The moment I see this, I naturally go to the number line. (z positive): ______________________________ 0 ____z_________________3z___________m____________________ OR (z negative): __3z_________(m)______________z___(m)__ 0 _______________(m)____________________ Both z and m could be positive, both negative or z negative m positive. Not sufficient. (2) 4zm > 0 4z > m (z positive): __________________(m)____________ 0 __(m)_______z_____________________(m)_____4z______ OR (z negative): ______m_______4z____________________z_____ 0 _______________________ Both z and m could be positive, both negative or m negative z positive. Not sufficient. Both together, 3z < m < 4z (z positive): ______________________________ 0 ____z_________________3z_____m_______4z___________ OR (z negative): ___________4z________3z___________________z_____ 0 _______________________ m has to be to the left of 4z but right of 3z. Not possible. So m and z both must be positive. So m + z > 0 Answer (C)
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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09 May 2016, 21:52
Thank you so much Karishma. You're a saviour!
Is it possible to visualise such problems rather than drawing on number line? I think that will be prone to errors.
Besides, I have noticed that problem statements concerning number line involves too many possibilities as you explained in detail. For the same reason, I feel these questions are designed to put the test taker in a situation where it consumes more than required time? My gut will always push me to rush on these questions therefore.
Rephrasing the question is vital I guess. Let me know if the only way through such questions is to draw the number line and analyze?
Cheers!
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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10 May 2016, 19:43
HarisinghKhedar wrote: Thank you so much Karishma. You're a saviour!
Is it possible to visualise such problems rather than drawing on number line? I think that will be prone to errors.
Besides, I have noticed that problem statements concerning number line involves too many possibilities as you explained in detail. For the same reason, I feel these questions are designed to put the test taker in a situation where it consumes more than required time? My gut will always push me to rush on these questions therefore.
Rephrasing the question is vital I guess. Let me know if the only way through such questions is to draw the number line and analyze?
Cheers!
Posted from my mobile device Most questions can be solved using many different methods. You can use the graphical approach discussed by walker here: http://gmatclub.com/forum/ismz75657.htmlAs for too many possibilities on the number line, it seems that way when I draw it out but here is how I see it in my mind. Stmtn 1: There are only 2 cases: z is positive (or 0) or z is negative. If z is to the right of 0, 3z is further to the right of z and m is further to the right of 3z. m has to be positive. If z is to the left of 0, 3z is further to the left of z but m is anywhere on the right of 3z. m could be negative or positive. That's all I need to know.
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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14 Oct 2016, 04:55
smitakokne wrote: Is m+z > 0
(1) m3z > 0 (2) 4zm > 0 We must determine whether the sum of m and z is greater than zero. Statement One Alone: m – 3z > 0 We can manipulate the inequality in statement one to read: m > 3z However, we still cannot determine whether the sum of m and z is greater than zero. Statement one alone is insufficient to answer the question. We can eliminate answer choices A and D. Statement Two Alone: 4z – m > 0 We can manipulate the inequality in statement two to read: 4z > m However, we still cannot determine whether the sum of m and z is greater than zero. Statement two alone is insufficient to answer the question. We can eliminate answer choice B. Statements One and Two Together: Using our statements together, we can add together our two inequalities. Rewriting statement 2 as –m + 4z > 0, we have m – 3z > 0 + (m +4z > 0) z > 0 Since we know that z > 0 and that m > 3z (from statement one), m must also be greater than zero. Thus, the sum of m and z is greater than zero. Answer: C
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Re: Is m + z > 0 (1) m  3z > 0 (2) 4z  m > 0
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03 May 2018, 19:57
Bunuel wrote: Is m+z > 0(1) m  3z > 0. Insufficient on its own. (2) 4z  m > 0. Insufficient on its own. (1)+(2) Remember we can add inequalities with the sign in the same direction: \((m3z)+(4zm)>0\); \(z>0\), so \(z\) is positive. From (1) \(m>(3z=positive)\), so \(m\) is positive too (\(m\) is more than some positive number \(3z\), so it's positive). Therefore, \(m+z=positive+positive>0\). Sufficient. Answer: C. For graphic approach refer to: http://gmatclub.com/forum/ismz01m ... 75657.html For these types of questions, is there a rule of thumb or a easy way to tell whether the statements alone will be sufficient? I always waste a lot of time on these types of questions.




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